Let $f$ be in $L^1_{loc}(\mathbb{R}^n)$. If for all $x,x_0$ and for all $r>0$ we have $$\frac{1}{m(B_r(x_0))}\int_{B_r(x_0)}|f(x)-f(x_0)|\,dx < C\cdot r^{\alpha} $$
where $C$ doesn't depend on $x,x_0$ or $r$. Then $f\in C^{0,\alpha}(\mathbb{R}^n)$.
One can see that the set of points for a given $x_0$ for which the Hölder condition doesn't hold is an open set. But this may not be enough to cause a contradiction in the bound of the integral in the RHS.
At first I doubted this was so, but in fact it's trivial. Since this smells like a homework problem, a sketch:
Say $r=|x-y|$. Note that $$|f(x)-f(y)|=\frac1{m(B_r(x))}\int_{B_r(x)}|f(x)-f(y)|\,dt.$$Now use the fact that $$|f(x)-f(y)|\le|f(x)-f(t)|+|f(y)-f(t)|$$and note that $B_r(x)\subset B_{2r}(y)$.